What is inverse oflaplace 2/((s^2+1)) ?

The answer to inverse oflaplace 2/((s^2+1)) is 2sin(t)

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inverse oflaplace 2/((s^2+1))

inverse laplace

\frac{2}{( s ^{2} + 1 )}

2 sin (t)

Solution steps

L^{- 1} {\frac{2}{( s ^{2} + 1 )}}

= L^{- 1} {2 \cdot \frac{1}{s ^{2} + 1 ^{2}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 2 L^{- 1} {\frac{1}{s ^{2} + 1 ^{2}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{a}{s ^{2} + a ^{2}}} = sin (a t)

L^{- 1} {\frac{1}{s ^{2} + 1 ^{2}}} = sin (t)

= 2 sin (t)

\int 6 csc (x) d x

\int cos (x) \cdot sin^{4} (x) d x

\int \frac{( x - 1 )}{x ^{2} - x + 1} d x

x \to - \infty lim (e^{x} - 1)

x \to \frac{π}{4} lim (\frac{sec ^{2} ( x ) - 2 tan ( x )}{1 + cos ( 4 x )})

What is inverse oflaplace 2/((s^2+1)) ?
The answer to inverse oflaplace 2/((s^2+1)) is 2sin(t)